Educational Codeforces Round 47

A. Game Shopping

签.

 #include <bits/stdc++.h>
 using namespace std;

 #define N 1010
 int n, m, c[N];
 queue <int> q;

 int main()
 {
     while (scanf("%d%d", &n, &m) != EOF)
     {
         while (!q.empty()) q.pop();
         for (int i = ; i <= n; ++i) scanf("%d", c + i);
         for (int i = , x; i <= m; ++i)
         {
             scanf("%d", &x);
             q.push(x);
         }
         int res = ;
         for (int i = ; i <= n && !q.empty(); ++i)
         {
             if (q.front() >= c[i])
             {
                 q.pop();
                 ++res;
             }
         }
         printf("%d\n", res);
     }
     return ;
 }

B. Minimum Ternary String

签.

 #include <bits/stdc++.h>
 using namespace std;

 #define N 100010
 char s[N];
 int n, cnt;

 int main()
 {
     while (scanf("%s", s + ) != EOF)
     {
         n = strlen(s + );
         cnt = ;
         for (int i = ; i <= n; ++i) if (s[i] == '')
             ++cnt;
         bool flag = false;
         for (int i = ; i <= n; ++i)
         {
             if (flag && s[i] != '')
                putchar(s[i]);
             else if (!flag)
             {
                 if (s[i] == '')
                     putchar(s[i]);
                 else if (s[i] == '')
                 {
                     flag = true;
                     while (cnt--) putchar('');
                     putchar(s[i]);
                 }
             }
         }
         if (cnt > ) while (cnt--) putchar('');
         puts("");
     }
     return ;
 }

C. Annoying Present

签.

 #include <bits/stdc++.h>
 using namespace std;

 #define db long double
 #define ll long long
 #define N 100010
 int n, m, x[N], d[N];

 ll f(int x)
 {
     return 1ll * n * (n + ) /  + (1ll * x * x - 1ll * n * x - x);
 }

 int main()
 {
     while (scanf("%d%d", &n, &m) != EOF)
     {
         ll Max = (ll)-1e18, Min = (ll)1e18;
         for (int i = ; i <= n; ++i)
         {
             ll tmp = f(i);
             Max = max(Max, tmp);
             Min = min(Min, tmp);
         }
         for (int i = ; i <= m; ++i)
             scanf("%d%d", x + i, d + i);
         db res = ;
         for (int i = ; i <= m; ++i)
         {
             res += 1ll * n * x[i];
             if (d[i] < )
                 res += 1ll * d[i] * Min;
             else
                 res += 1ll * d[i] * Max;
         }
         printf("%.10Lf\n", res * 1.0 / n);
     }
     return ;
 }

D. Relatively Prime Graph

签。

$考虑质数的密度很大 并且\phi(p) = p - 1 所以暴力找边即可$

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 #define N 100010
 #define pii pair <int, int>
 int n, m;
 vector <pii> res;
 int phi[N], prime[N];
 bool check[N];
 int tot;

 void init()
 {
     memset(check, false, sizeof check);
     phi[] = ;
     tot = ;
     for (int i = ; i < N; ++i)
     {
         if (!check[i])
         {
             prime[++tot] = i;
             phi[i] = i - ;
         }
         for (int j = ; j <= tot; ++j)
         {
             if (i * prime[j] >= N) break;
             check[i * prime[j]] = true;
             if (i % prime[j] == )
             {
                 phi[i * prime[j]] = phi[i] * prime[j];
                 break;
             }
             else
                 phi[i * prime[j]] = phi[i] * (prime[j] - );
         }
     }
 }

 int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

 int main()
 {
     init();
     while (scanf("%d%d", &n, &m) != EOF)
     {
         ll sum = ;
         for (int i = ; i <= n; ++i) sum += phi[i];
         if (m < n -  || m > sum)
         {
             puts("Impossible");
             continue;
         }
         puts("Possible");
         for (int i = ; i <= n; ++i) printf("%d %d\n", , i);
         m -= n - ;
         vector <int> vec;
         for (int i = ; i <= n; ++i) vec.push_back(i);
         sort(vec.begin(), vec.end(), [](int a, int b) { return phi[a] > phi[b]; });
         for (auto it : vec)
         {
             for (int i = ; i < it && m > ; ++i) if (gcd(it, i) == )
             {
                 printf("%d %d\n", it, i);
                 --m;
             }
             if (m <= ) break;
         }
     }
     return ;
 }

E. Intercity Travelling

签。

统计$a_i$被计算多少次

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 #define N 1000010
 const ll p = (ll);
 int n;
 ll a[N], Bit[N];

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%lld", a + i);
         Bit[] = ;
         for (int i = ; i <= n; ++i) Bit[i] = (Bit[i - ] * ) % p;
         if (n == )
         {
             printf("%lld\n", a[]);
             continue;
         }
         ll res = ;
         for (int i = ; i <= n; ++i)
         {
             res = (res + Bit[n - i] * a[i] % p) % p;
             if (n - i -  >= )
                 res = (res + 1ll * (n - i) * Bit[n - i - ] % p * a[i] % p) % p;
         }
         printf("%lld\n", res);
     }
     return ;
 }